The following inequality is given in some of Yale's online lecture notes

$$P(|Z|>x) \leq 2 \sqrt{2 \pi} \phi(x)$$

Where $Z \sim N(0,1)$ with density $\phi(x)$. They call it a concentration inequality, the proof is not given.

I know how to prove the bounds

$$(1/x - 1/x^3)\phi(x) \leq 1-\Phi(x) \leq 1/x \phi(x)$$

But the method I used for these did not get me anywhere.

I know that it is enough to show

$$\Phi(x) \geq 1- \sqrt{2 \pi} \phi(x)$$

by symmetry of normal distribution. So we need to show

$$\int_x^\infty \phi(y)dy \geq 1- \sqrt{2 \pi} \phi(x)$$


Following advice from answer

Note that $E[e^{Zt}] = e^{t^2/2}$ and we already have the inequality

$$P(Z>x) \leq E[e^{Zt}]/e^{xt}$$

Maximising the RHS equality with respect to $t$ gives $t_\max = x$, and gives the upperbound needed.


1 Answer 1


It is not true. When $x=0.1$, $P(|Z|>x) = 0.92$ and $2\phi(x)=0.79$.

Maybe the correct one is:

$$P(|Z|>x) \leq 2 \sqrt{2\pi}\phi(x)$$

For new inequality, following follow step:

  1. Get $E(e^{\lambda Z})$

  2. Find $\sup_\lambda(\lambda x - \log E(e^{\lambda Z}))$

  3. Use Chernoff bound, you can get $\Pr(Z>x) < e^{-t^2/2}$

  • $\begingroup$ You're right. I forgot the normalisingn constant is not included. Will correct my post $\endgroup$
    – Xiaomi
    Dec 1, 2018 at 7:46
  • $\begingroup$ Thanks a lot! Maximising the bound never occured to me but it's so obvious now :) $\endgroup$
    – Xiaomi
    Dec 1, 2018 at 11:21

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